Question

Prove lemma 10.27.

Short answer

Lemma 10.27 is proved.

Step-by-step solution

Statement of lemma 10.27

If R is an Integral Domain and F is as above, then for all non-zero $a,b,c,d,k\in R$:

(i) role="math" localid="1653709959557" $\left[0_R,b\right]=\left[0_R,d\right];$

(ii) $\left[a,b\right]=\left[ak,bk\right];$

(iii) $\left[a,a\right]=\left[c,c\right].$

Show that 0R,b=0R,d

Let R be an integral domain.

Let F denote the set of all equivalence classes under $~$.

Therefore, by definition of $~\left[a,b\right]=\left[c,d\right]$ in F if and only if$ad=bc$ in R.

Now, let$a=c=0_R$ .

Therefore, it can be written as:

$\begin{array}{rcl}{0}_{R}b& =& b{0}_R\\ 0& =& 0\\ ad& =& bc\end{array}$

This implies,$\left[0_R,b\right]=\left[0_R,d\right]$ .

Hence, it is proved$\left[0_R,b\right]=\left[0_R,d\right]$ .

Show that a,b=ak,bk

Let R be an integral domain.

Let F denote the set of all equivalence classes under $~$.

Therefore, by definition of $~$, $\left[a,b\right]=\left[c,d\right]$in F if and only if $ad=bc$in R.

Now, let$c=ak,d=bka=a,b=b$ .

$$\begin{gathered} \Rightarrow ad=abk \\ \Rightarrow bc=bak=abk\left[\therefore Risanintegraldomain\right] \\ \Rightarrow ad=bc \\ \Rightarrow \left[a,b\right]=\left[c,d\right] \end{gathered}$$

This implies, $\left[a,b\right]=\left[ak,bk\right]$.

Hence, it is proved that$\left[a,b\right]=\left[ak,bk\right]$ .

Show that a,a=c,c

Let R be an integral domain.

Let F denote the set of all equivalence classes under $~$.

Therefore, by definition of $~$, $\left[a,b\right]=\left[c,d\right]$in F if and only if $ad=bc$in R.

Let $a=b=a$and $c=d=c$.

Then,$ad=ac$and$bc=ac$ .

Simplify$ad=ac$ as:

$\begin{array}{rcl}ad& =& bc\\ \left[a,b\right]& =& \left[c,d\right]\\ \left[a,a\right]& =& \left[c,c\right]\end{array}$

Hence, it is proved$\left[a,a\right]=\left[c,c\right]$ .